Latest TCS Aptitude Question SOLUTION: A drawer holds 4 red hats and 4 blue hats. what is probability of getting exactly three blue hats when taking out 4 hats randomly out of the drawer and immediately returning every

• Placement Papers
• /
• TCS
• /
• Numerical Ability
• /
• Probability

TCS Question Papers: TCS Interview Experience(90) TCS Company Profile

(#M40010725) TCS Numerical Ability question Keep an EYE

A drawer holds 4 red hats and 4 blue hats. what is probability of getting exactly three blue hats when taking out 4 hats randomly out of the drawer and immediately returning every has to the drawer before taking out the next?
a)1/2
b)1/8
c)1/4
d)3/8

Asked In TCS krishnaprasad TATA (11 years ago)

Unsolved Read Solution (6)

Is this Puzzle helpful?   (8)   (6) Submit Your Solution Probability

Read Solution (6) :

krishnamurthy  (11 years ago)

Correct answer is option (a)1/2
1/4+1/4=1/2

Is this solution Helpfull? Yes (14) | No (12)  

Garima  (11 years ago)

1/4

out of possible 16 outcomes, 4 are desired ones in which we are getting exactly three blue hats when taking out 4 hats randomly out of the drawer and immediately returning every has to the drawer before taking out the next.
these combinations are
BBBR
BBRB
BRBB
RBBB

so prob = 4/16 =1/4

Is this solution Helpfull? Yes (13) | No (15)  

Bhargav A patel  (11 years ago)

I m not sure about my solution.... if right plz give feed back...
here we have total 8 hat so its total combination is = 8c4
and here three blue hat have to take so remaining 1 hat can be red ..
so for blue hat =4c3
red hat=4c1
probability=event occurance/total possible value
probability=4c3*4c1/8c4
s0 approximate ans is 1/4

Is this solution Helpfull? Yes (6) | No (16)  

Vivek Ausekar  (11 years ago)

there are tottal 8 hats
we will totaly select 4 from 8
and 3 from 4 blue hats
and 1 from 4 red hat
so as per product rule
(3/4*1/4)/(4/8)=3/4*1/4*2=3/4*1/2=3/8

Is this solution Helpfull? Yes (3) | No (11)  

Nidhi Patel  (11 years ago)

@KRISHNAMURTHY ,, kindly explain how is it 1/2 ?

Is this solution Helpfull? Yes (3) | No (2)  

abhishek pratap singh  (11 years ago)

let red hat is success , prob of success is 4/8 = 1/2
for first case 3 red hat - 4C3(1/2)^3(1/2) since nCr(succ)^r(failure)^(n-r)
failure= 1- succ
similarly for second case 4C3(1/2)^3(1/2)
add both case 2*4C3(1/2)^3(1/2) = 2*4*1/16 =1/2

Is this solution Helpfull? Yes (1) | No (6)  

Please Login to Read Solution.